@typeswitch
tl;dr solution at bottom

I guess we need to clarify what the allowed operation is.
e.g. if I am allowed to whether any (natural?) linear combination of outcomes is probability 1/2, then I can just ask is 3*P(rolling i) = 1/2.

If the question I'm allowed to ask is
P(rolling i, j, or k) = 1/2, then we are essentially asking for an invertible 6x6 matrix with one row of 1s, and five rows containing a permutation of [1, 1, 1, 0, 0, 0]. In which case, again you are using 5 tests

That is certainly brute forceable by computer (at most (6 choose 3) choose 5 = 15,504 meaningfully different possibilities), but there may be a clever rank argument that it's possible or not.

Okay, I did some more thinking while doing my laundry, and here's a strategy:

If a + b + c = 0.5 and d + b + c = 0.5, then a = d. Repeat with modifications four more times to show that that all are equal to one another.

The problem with that is that doing it naively requires six tests, so seven equations, if we include that they all sum to one, my bad. So you need to take advantage of all the information you've gathered in the first four iterations and choose smartly in the last one.

Here is a solution written as a matrix equation

\(\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 \\
1 & 1 & 1 & 0 & 0 & 0\\
1 & 1 & 0 & 1 & 0 & 0 \\
1 & 1 & 0 & 0 & 1 & 0\\
1 & 1 & 0 & 0 & 0 & 1 \\
0 & 1 & 1 & 1 & 0 & 0\\
\end{pmatrix}
\cdot \begin{pmatrix}
1/6 \\ 1/6 \\ 1/6 \\ 1/6 \\ 1/6 \\ 1/6
\end{pmatrix}
= \begin{pmatrix}
1 \\ 1/2 \\ 1/2 \\ 1/2 \\ 1/2 \\ 1/2
\end{pmatrix}
\)

One final comment, because I can't help myself. It shouldn't be too hard to generalise this pattern to other *even* sizes of dice.

#linearalgebra #probability